Is there a multi-dimensional version of arange/linspace in numpy?

Question

I would like a list of 2d NumPy arrays (x,y) , where each x is in {-5, -4.5, -4, -3.5, ..., 3.5, 4, 4.5, 5} and the same for y.

I could do

x = np.ar         


        

Answer 1

You can use np.mgrid for this, it's often more convenient than np.meshgrid because it creates the arrays in one step:

import numpy as np
X,Y = np.mgrid[-5:5.1:0.5, -5:5.1:0.5]

For linspace-like functionality, replace the step (i.e. 0.5) with a complex number whose magnitude specifies the number of points you want in the series. Using this syntax, the same arrays as above are specified as:

X, Y = np.mgrid[-5:5:21j, -5:5:21j]

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You can then create your pairs as:

xy = np.vstack((X.flatten(), Y.flatten())).T

As @ali_m suggested, this can all be done in one line:

xy = np.mgrid[-5:5.1:0.5, -5:5.1:0.5].reshape(2,-1).T

Best of luck!

Answer 2

I think you want np.meshgrid:

Return coordinate matrices from coordinate vectors.

Make N-D coordinate arrays for vectorized evaluations of N-D scalar/vector fields over N-D grids, given one-dimensional coordinate arrays x1, x2,..., xn.

import numpy as np
x = np.arange(-5, 5.1, 0.5)
y = np.arange(-5, 5.1, 0.5)
X,Y = np.meshgrid(x,y)

you can convert that to your desired output with

XY=np.array([X.flatten(),Y.flatten()]).T

print XY
array([[-5. , -5. ],
       [-4.5, -5. ],
       [-4. , -5. ],
       [-3.5, -5. ],
       [-3. , -5. ],
       [-2.5, -5. ],
       ....
       [ 3. ,  5. ],
       [ 3.5,  5. ],
       [ 4. ,  5. ],
       [ 4.5,  5. ],
       [ 5. ,  5. ]])

Answer 3

Based on this example, you can make any dim you want

def linspace3D(point1,point2,length):
    v1 = np.linspace(point1[0],point2[0],length)
    v2 = np.linspace(point1[1],point2[1],length)
    v3 = np.linspace(point1[2],point2[2],length)
    line = np.zeros(shape=[length,3])
    line[:,0]=v1
    line[:,1]=v2
    line[:,2]=v3
    return line

Answer 4

I still did it with Linspace because I prefer to stick to this command.

You can create like the following format: np.linspace(np.zeros(width)[0], np.full((1,width),-1)[0], height)

np.linspace(np.zeros(5)[0],np.full((1,5),-1)[0],5)

Output the following:

array([[ 0.  ,  0.  ,  0.  ,  0.  ,  0.  ],
       [-0.25, -0.25, -0.25, -0.25, -0.25],
       [-0.5 , -0.5 , -0.5 , -0.5 , -0.5 ],
       [-0.75, -0.75, -0.75, -0.75, -0.75],
       [-1.  , -1.  , -1.  , -1.  , -1.  ]])

Add .tranpose() then you get:

array([[ 0.  , -0.25, -0.5 , -0.75, -1.  ],
      [ 0.  , -0.25, -0.5 , -0.75, -1.  ],
      [ 0.  , -0.25, -0.5 , -0.75, -1.  ],
      [ 0.  , -0.25, -0.5 , -0.75, -1.  ],
      [ 0.  , -0.25, -0.5 , -0.75, -1.  ]])

Answer 5

Not sure if I understand the question - to make a list of 2-element NumPy arrays, this works:

import numpy as np
x = np.arange(-5, 5.1, 0.5)
X, Y = np.meshgrid(x, x)
Liszt = [np.array(thing) for thing in zip(X.flatten(), Y.flatten())] # for python 2.7

zip gives you a list of tuples, and the list comprehension does the rest.

Answer 6

It is not super fast solution, but works for any dimension

import numpy as np
def linspace_md(v_min,v_max,dim,num):
    output = np.empty( (num**dim,dim)  )
    values = np.linspace(v_min,v_max,num)
    for i in range(output.shape[0]):
        for d in range(dim):
            output[i][d] = values[( i//(dim**d) )%num]
    return output